Question

How do you find the vertical, horizontal and slant asymptotes of: `(x^2-25)/(x^2+5x)`?

Answer 1

vertical asymptote: `x=0`
horizontal asymptote: `f(x)=1`
slant asymptote: does not exist

Explanation:

Finding the Vertical Asymptote

Given,

`f(x)=(x^2-25)/(x^2+5x)`

Factor the numerator.

`f(x)=((x+5)(x-5))/(x(x+5)`

Cancel out any factors that appear in the numerator and denominator.

`f(x)=(x-5)/x`

Set the denominator equal to `0` and solve for `x`.

`color(green)(|bar(ul(color(white)(a/a)color(black)(x=0)color(white)(a/a)|)))`

Finding the Horizontal Asymptote

Given,

`f(x)=(color(darkorange)1x^2-25)/(color(purple)1x^2+5x)`

Divide the `color(darkorange)("leading coefficient")` of the leading term in the numerator by the `color(purple)("leading coefficient")` of the leading term in the denominator.

`f(x)=color(darkorange)1/color(purple)1`

`color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=1)color(white)(a/a)|)))`

Finding the Slant Asymptote

Given,

`f(x)=(x^2-25)/(x^2+5x)`

There would be a slant asymptote if the degree of the leading term in the numerator was `1` value larger than the degree of the leading term in the denominator. In your case, we see that the degree in both the numerator and denominator are equal.

`:.`, the slant asymptote does not exist.